Chapter 7 ENTROPY Dr Ali Jawarneh Department of Mechanical Engineering Hashemite University
Objectives Apply the second law of thermodynamics to processes. Define a new property called entropy to quantify the secondlaw effects. Establish the increase of entropy principle. Calculate the entropy changes that take place during processes for pure substances, incompressible substances, and ideal gases. Examine a special class of idealized processes, called isentropic processes, and develop the property relations for these processes. Derive the reversible steady-flow work relations. Develop the isentropic efficiencies for various steady-flow devices. Introduce and apply the entropy balance to various systems. 2
7-1 ENTROPY which is a quantitative measure of microscopic disorder for a system. To demonstrate the validity of the Clausius inequality: Applying the energy balance to the combined system identified by dashed lines yields: let the system undergo a cycle while the cyclic device undergoes an integral number of cycles It appears that the combined system is exchanging heat with a single thermal energy reservoir while involving i (producing or consuming) work W C during a cycle. On the basis of the Kelvin Planck statement of the second law, which states that no system can produce a net amount of work while operating in a cycle and exchanging heat with a single thermal energy reservoir, we reason that W C cannot be a work output, and thus it cannot be a positive quantity. The system considered in the development of the Clausius inequality. T (variable) Clasius inequality This inequality is valid for all cycles, reversible or irreversible, including the refrigeration cycles. The symbol (integral symbol with a circle in the middle) is used to indicate that the integration is to be performed over the entire cycle. Any y heat transfer to or from a system can be considered to consist of differential amounts of heat transfer. Then the cyclic integral of δq/t can be viewed as the sum of all these differential amounts of heat transfer divided by the temperature at the boundary. 3
If no irreversibilities occur within the system as well as the reversible cyclic device, then the cycle undergone by the combined system is internally reversible. As such, it can be reversed. In the reversed cycle case, all the quantities have the same magnitude but the opposite sign. Therefore, the work W C, which could not be a positive quantity in the regular case, cannot be a negative quantity in the reversed case. Then it follows that since it cannot be a positive or negative quantity, and therefore For internally reversible cycles. Thus, we conclude that the equality in the Clausius inequality holds for totally or just internally reversible cycles and the inequality for the irreversible ones. a quantity whose cyclic integral is zero depends on the state only and not the process path, and thus it is a property. Therefore, the quantity (δq/t ) int rev must represent a property in the differential form. Clausius realized in 1865 that he had discovered a new thermodynamic property, and he chose to name this property entropy. It is designated S and is defined as Formal definition of entropy Entropy is an extensive property of a system and sometimes is referred to as total entropy. Entropy per unit mass, designated s,, is an intensive property p and has the unit kj/kg K. The entropy change of a system during a process can be determined by integrating Eq. 7 4 between the initial and the final states: 4
A quantity whose cyclic integral is zero (i.e., a property like volume) Entropy is an extensive property of a system. The net change in volume (a The entropy change between two property) during specified states is the same whether a cycle is the process is reversible or irreversible. always zero. To perform the integration in Eq. 7 5, one needs to know the relation between Q and T during a process. This relation is often not available, and the integral in Eq. 7 5 can be performed for a few cases only. For the majority of cases we have to rely on tabulated data for entropy. Note that entropy is a property, and like all other properties, it has fixed values at fixed states. Therefore, the entropy change ΔS between two specified states is the same no matter what path, reversible or irreversible, is followed during a process (Fig. 7 3). Also note that the integral of δq/t gives us the value of entropy change only if the integration is carried out along an internally reversible path between the two states. The integral of δq/t along an irreversible path is not a property, and in general, different values will be obtained when the integration is carried out along different irreversible paths. Therefore, even for irreversible processes, the entropy change should be determined by carrying out this integration along some convenient imaginary internally reversible path between the specified states. 5
A Special Case: Internally Reversible Isothermal Heat Transfer Processes Recall that isothermal heat transfer processes are internally reversible. This equation is particularly useful for determining the entropy changes of thermal energy reservoirs. Notice that the entropy change of a system during an internally reversible isothermal process can be positive or negative, depending on the direction of heat transfer. Heat transfer to a system increases the entropy of a system, whereas heat transfer from a system decreases it. In fact, losing heat is the only way the entropy of a system can be decreased. 6
7-2 THE INCREASE OF ENTROPY PRINCIPLE The equality holds for an internally reversible process and the inequality for an irreversible process. A cycle composed of a reversible and an irreversible process. entropy change of the system entropy generation due to irreversiblities entropy transfer with heat Some entropy is generated or created during an irreversible process, and this generation is due entirely to the presence of irreversibilities. The entropy generation S gen is always a positive quantity or zero. Its value depends on the process, and thus it is not a property of the system. Heat transfer to a system increases the entropy, and heat transfer from a system decreases it. The effect of irreversibilities is always to increase the entropy. 7
A system and its surroundings, for example, constitute an isolated system since both can be enclosed by a sufficiently large arbitrary boundary across which there is no heat, work, or mass transfer. Therefore, a system and its surroundings can be viewed as the two subsystems of an isolated system The entropy change of an isolated system is the sum of the entropy changes of its components, and is never less than zero. Isolated system (or simply an adiabatic closed system) The increase of entropy principle A system and its surroundings form an isolated system. 8
Some Remarks about Entropy The entropy change of a system can be negative, but the entropy generation cannot. 1. Processes can occur in a certain direction only, not in any direction. A process must proceed in the direction that complies with the increase of entropy principle, that is, S gen 0. A process that violates this principle is impossible. 2. Entropy is a nonconserved property, p and there is no such thing as the conservation of entropy principle. Entropy is conserved during the idealized reversible processes only and increases during all actual processes. 3. The performance of engineering systems is degraded by the presence of irreversibilities, and entropy generation is a measure of the magnitudes of the irreversibilities during that process. It is also used to establish criteria for the performance of engineering devices. 9
Example: Air is compressed by a 12-kW compressor from P 1 to P 2. The air temperature is maintained constant at 25 C during this process as a result of heat transfer to the surrounding medium at 10 C. Determine the rate of entropy change of the air. State the assumptions made in solving this problem. 10
Solution: Assumptions 1- This is a steady-flow process since there is no change with time. 2- Kinetic and potential energy changes are negligible. 3 - Air is an ideal gas. 4 - The process involves no internal irreversibilities such as friction, and thus it is an isothermal, internally reversible process. Properties Noting that h = h(t) for ideal gases, we have h 1 = h 2 since T 1 = T 2 = 25 C. Noting that the process is assumed to be an isothermal and internally reversible process, the rate of entropy change of air is determined to be The negative sign indicates that the entropy of the system is decreasing during this process. This is not a violation of the second law, however, since it is the entropy generation S gen that cannot be negative. 11
Example: A completely reversible air conditioner provides 36,000 kj/h of cooling for a space maintained at 20 0 C while rejecting heat to the environmental air at 45 0 C. Calculate the rate at which the entropies of the two reservoirs change and verify that this air conditioner satisfies the increase of entropy principle. i 12
Solution: The net rate of entropy change is zero as it must be in order to satisfy the second law. 13
7-3 ENTROPY CHANGE OF PURE SUBSTANCES Entropy is a property, and thus the value of entropy of a system is fixed once the state of the system is fixed. Schematic of the T-s diagram for water. The entropy of a pure substance is determined from the tables (like other properties). Entropy change 14
7-4 ISENTROPIC PROCESSES A process during which the entropy remains constant is called an isentropic process(internally reversible and adiabatic). Isentropic process(valid for closed system or control volume) During an internally reversible, adiabatic (isentropic) i process, the entropy remains constant. isentropic processes enable us to define efficiencies for processes to compare the actual performance of these devices to the performance under idealized conditions. The isentropic process appears as a vertical line segment on a T-s diagram. 15
Example: Refrigerant-134a enters an adiabatic compressor as saturated vapor at 160 kpa at a rate of 2 m 3 /min and is compressed to a pressure of 900 kpa. Determine the minimum power that must be supplied to the compressor. 16
Solution: The power input to an adiabatic compressor will be a minimum when the compression process is reversible. (adiabatic + reversible=isentropic) i From the refrigerant tables (Tables A-11 through A-13), 17
Example: An insulated piston cylinder device contains 5 L of saturated liquid water at a constant pressure of 150 kpa. An electric resistance heater inside the cylinder is now turned on, and 2200 kj of energy is transferred to the steam. Determine the entropy change of the water during this process. 18
Solution: From the steam tables (Tables A-4 through A-6), 19
7-5 PROPERTY DIAGRAMS INVOLVING ENTROPY internally reversible isothermal process: On a T-S diagram, the area under the process curve represents the heat transfer for internally reversible processes. For adiabatic steady-flow devices such as turbines, compressors, and nozzles. The vertical distance h on an h-s diagram is a measure of work, and the horizontal distance s is a measure of irreversibilities. Mollier diagram: The h-s diagram 20
Mollier diagram: The h-s diagram 21
Carnot cycle 22
Example: Refrigerant-134a at 240 kpa and 20 0 C undergoes an isothermal process in a closed system until its quality is 20 percent. On per unit mass basis, determine how much work and heat transfer are required. 23
Solution: (Table A-11) The negative sign shows that the heat is actually transferred from the system. 24
Example: Determine the total heat transfer for the reversible process 1-3. Solution: 25
7-6 WHAT IS ENTROPY? Boltzmann relation which is a quantitative measure of microscopic disorder for a system k:boltzmann constant p: thermodynamic probability, molecular randomness or uncertainty t (i.e., molecular l probability) A pure crystalline substance at absolute zero temperature is in perfect order, and its entropy is zero (the third law of thermodynamics). The level of molecular disorder (entropy) of a substance increases as it melts or evaporates. Disorganized energy does not create much useful effect, no matter how large it is. Molecules in the gas phase possess a considerable amount of kinetic energy. However, we know that no matter how large their kinetic energies are, the gas molecules do not rotate a paddle wheel inserted into the container and produce work. This is because the gas molecules, and the energy they possess, are disorganized. Probably the number of molecules trying to rotate the wheel in one direction at any instant is equal to the number of molecules that are trying to rotate it in the opposite direction, causing the wheel to remain motionless. Therefore, we cannot extract any 26 useful work directly from disorganized energy
In the absence of friction, raising a weight by a rotating shaft does not create any disorder (entropy), and thus energy is not degraded during this process. Now consider a rotating ti shaft. This time the energy of the molecules is completely organized since the molecules of the shaft are rotating in the same direction together. This organized energy can readily be used to perform useful tasks such as raising a weight or generating electricity. Being an organized form of energy, work is free of disorder or randomness and thus free of entropy. There is no entropy transfer associated with energy transfer as work. Therefore, in the absence of any friction, the process of raising a weight by a rotating shaft (or a flywheel) does not produce any entropy. Any process that does not produce a net entropy is reversible, and thus the process just described can be reversed by lowering the weight. Therefore, energy is not degraded during this process, and no potential to do work is lost. 27
Instead of raising a weight, let us operate the paddle wheel in a container filled with a gas.the paddle-wheel work in this case is converted to the internal energy of the gas, as evidenced by a rise in gas temperature, creating a higher level of molecular disorder in the container. This process is quite different from raising a weight since the organized paddle-wheel energy is now converted to a highly hl disorganized i d form of The paddle-wheel work done on energy, which cannot be converted back to a gas increases the level of the paddle wheel as the rotational kinetic disorder (entropy) of the gas, energy. Only a portion of this energy can be converted to work by partially reorganizing it through the use of a heat engine. Therefore, energy is degraded during this process, the ability to do work is reduced, molecular disorder is produced, and associated with all this is an increase in entropy. and dthus energy is degraded d d during this process. 28
The quantity of energy is always preserved during an actual process (the first law), but the quality is bound to decrease (the second law). This decrease in quality is always accompanied by an increase in entropy. As an example, consider the transfer of 10 kj of energy as heat from a hot medium to a cold one. At the end of the process, we still have the 10 kj of energy, but at a lower temperature and thus at a lower quality. Heat is, in essence, a form of disorganized energy, and some disorganization (entropy) flows with heat. As a result, the entropy and the level of molecular disorder or randomness of the hot body decreases with the entropy and the level of molecular disorder of the cold body increases. The second law requires that the increase in entropy of the cold body be greater than the decrease in entropy of the hot body, and thus the net entropy of the combined system (the cold body and the hot body) increases. That is, the combined system is at a state of greater disorder at the final state. Thus we can conclude that processes can occur only in the direction of increased overall entropy or molecular disorder. That is, the entire universe is getting more and more chaotic every day. During a heat transfer process, the net entropy increases. (The increase in the entropy of the cold body more than offsets the decrease in the entropy of the hot body.) 29
Entropy can be viewed as a measure of molecular disorder, or molecular randomness. As a system becomes more disordered, the positions of the molecules l become less predictable and the entropy increases. Thus, it is not surprising that the entropy of a substance is lowest in the solid phase and highest in the gas phase In the solid phase, the molecules of a substance continually oscillate about their equilibrium positions, but they cannot move relative to each other, and their position at any instant can be predicted with good certainty. In the gas phase, however, the molecules move about at random, collide with each other, and change direction, making it extremely difficult to predict accurately the microscopic state of a system at any instant. Associated with this molecular chaos is a high value of entropy. The third law of thermodynamics the entropy of a pure crystalline substance at absolute zero temperature is zero provides an absolute reference point for the determination of entropy. The entropy determined relative to this point is called absolute entropy 30
7-7 THE T ds RELATIONS Closed stationary system The only type of work interaction a simple compressible system may involve as it undergoes an internally reversible process is the boundary work. the first Td ds, or Gibbs equation The Tdsrelations are valid for both reversible and irreversible processes and for both closed and open systems. These Td ds relations are developed dwith an internally reversible process in mind since the entropy change between two states must be evaluated along a reversible path. However, the results obtained are valid for both reversible and irreversible processes since entropy is a property and the change in a property between two states is independent of the type of process the system undergoes the second T ds equation Differential changes in entropy in terms of other properties 31
7-8 ENTROPY CHANGE OF LIQUIDS AND SOLIDS Since for liquids and solids Liquids and solids can be approximated as incompressible substances since their specific volumes remain nearly constant during a process. For and isentropic process of an incompressible substance Therefore, the isentropic process of an incompressible substance is also isothermal. 32
Example: A 25-kg iron block initially at 350 C is quenched in an insulated tank that contains 100 kg of water at 18 C. Assuming the water that t vaporizes during the process condenses back in the tank, determine the total entropy change during this process. Solution: Assumptions 1 Both the water and the iron block are incompressible substances with constant specific heats at room temperature. 2 The system is stationary and thus the kinetic and potential ti energies are negligible. ibl 3 The tank is well-insulated l and thus there is no heat transfer. 4 The water that evaporates, condenses back. Properties The specific heat of water at 25 C is c p = 4.18 kj/kg. C. The specific heat of iron at room temperature is c p = 0.45 kj/kg. C (Table A-3) 3). 33
closed system The entropy generated during this process is determined from Discussion: The results can be improved somewhat by using specific heats at average temperature. 34
Example: A 50-kg iron block and a 20-kg copper block, both initially at 80 C C, are dropped into a large lake at 15 C C. Thermal equilibrium is established after a while as a result of heat transfer between the blocks and the lake water. Determine the total entropy change for this process. Solution: Properties The specific heats of iron and copper at room temperature are c iron = 0.45 kj/kg. C and c copper = 0.386 kj/kg. C (Table A-3). Analysis The thermal-energy capacity of the lake is very large, and thus the temperatures of both the iron and the copper blocks will drop to the lake temperature (15 C) when the thermal equilibrium is established. Then the entropy changes of the blocks become 35
We take both the iron and the copper blocks, as the system. This is a closed system since no mass crosses the system boundary during the process. The energy balance for this system can be expressed as Then the total entropy change for this process is 36
7-9 THE ENTROPY CHANGE OF IDEAL GASES From the first T ds relation From the second T ds relation A broadcast from channel IG. The specific heats of ideal gases, with the exception of monatomic gases(such as helium), depend on temperature 37
Constant Specific Heats (Approximate Analysis) Table A-2a and A-2b Entropy change of an ideal gas on a unit mole basis Under the constant-specificheat assumption, the specific heat is assumed to be constant t at some average value. Δs = Δs M 38
Variable Specific Heats (Exact Analysis) When the temperature change during a process is large and the specific heats of the ideal gas vary nonlinearly within the temperature range, the assumption of constant specific heats may lead to considerable errors We choose absolute zero as the reference temperature and define a function s as Obviously, s is a function of temperature alone, and its value is zero at absolute zero temperature. On a unit mass basis Table A-17 to A-25 On a unit mole basis Note that unlike internal energy and enthalpy, the entropy of an ideal gas varies with specific volume or pressure as well as the temperature. Therefore, entropy cannot be tabulated as a function of temperature alone. The s values in the tables account for the temperature dependence of entropy. The variation of entropy with pressure is accounted for by the last term in Eq. 7 39. The entropy of an ideal gas depends on both T and P. The function s represents only the temperaturedependent part of entropy. 39
Example: 57 C 17 C Variable ab Specific c Heats Table A 17 Constant Specific Heats using a c p value at the average temperature of 37 C (Table A 2b) For small temperature differences, the exact and approximate relations for entropy changes of ideal gases give almost identical results. 40
Example: An insulated piston cylinder device initially contains 300 Lofairat120 at kpa and 17 C C. Air is now heated for 15 min by a 200-W resistance heater placed inside the cylinder. The pressure of air is maintained constant during this process. Determine the entropy change of air, assuming (a) constant specific heats and (b) variable specific heats. 41
Solution: The gas constant of air is R = 0.287 kj/kg.k (Table A-1). The energy balance for this stationary closed system since ΔU + W b = ΔH during a constant pressure quasi-equilibrium process. (a) Using a constant c p =1.02 kj/kg.k value from Table A-2b at the anticipated average temperature t of 450 K 42
(b) Assuming variable specific heats, From the air table (Table A-17): h 1 =290.16 kj/kg and @T 1 =290 K and =1.66802 kj/kg.k 43
Isentropic Processes of Ideal Gases Constant Specific Heats (Approximate Analysis) Setting this eq. equal to zero, we get The specific heat ratio k, in general, varies with temperature, and thus an average k value for the given temperature range should be used. we should refine the calculations{repeat the calculations} The isentropic relations of ideal gases are valid for the isentropic processes of ideal gases only. Equations 7 42 through h 7 44 can also be expressed in a compact form as 44
Isentropic Processes of Ideal Gases Variable Specific Heats (Exact Analysis) Relative Pressure and Relative Specific Volume exp(s /R) is the relative pressure P r. The use of P r data for calculating the final temperature during an isentropic process. T/P r is the relative specific volume v r. Table A 17 [Air] The use of v r data for calculating the final temperature during an isentropic process 45
Example: Air is compressed in a piston cylinder device from 100 kpa and 17 C to800 kpa in a reversible, adiabatic process. Determine the final temperature and the work done during this process, assuming (a) constant specific heats and (b) variable specific heats for air. 46
Solution: The gas constant of air is R = 0.287 kj/kg.k (Table A-1). The specific heat ratio of air at low to moderately high temperatures is k = 1.4 (Table A-2). (a) Assuming constant specific heats, the ideal gas isentropic relations give 47
(b) Assuming variable specific heats, the final temperature can be determined using the relative pressure data (Table A-17), 48
Example: Nitrogen at 120 kpa and 30 0 C is compressed to 600 kpa in an adiabatic compressor. Calculate l the minimum work needed for this process. Assume constant specific heats. 49
Solution: The properties p of nitrogen at an anticipated average temperature of 400 K are c p = 1.044 kj/kg K and k = 1.397 (Table A-2b). For the minimum work input to the compressor, the process must be reversible as well as adiabatic (i.e., isentropic). 50
Example: An ideal gas at 100 kpa and 27 C enters a steady-flow compressor. The gas is compressed to 400 kpa, and 10 percent of fthe mass that tentered dthe compressor is removed dfor some other use. The remaining 90 percent of the inlet gas is compressed to 600 kpa before leaving the compressor. The entire compression process is assumed to be reversible and adiabatic. The power supplied to the compressor is measured to be 32 kw. If the ideal gas has constant specific heats such that c v =0.8 kj/kg.k and c p = 1.1 kj/kg.k, (a) sketch the compression process on a T-s diagram, (b) determine the temperature of the gas at the two compressor exits, in K, and (c) determine the mass flow rate of the gas into the compressor, in kg/s. 51
Solution: 52
7-10 REVERSIBLE STEADY-FLOW WORK Recall that reversible (quasi-equilibrium) moving boundary work associated with closed systems is expressed in terms of the fluid properties as: Turbines We mentioned that the quasi-equilibrium work interactions lead to the maximum work output for work-producing devices and the minimum work When kinetic and potential energies are negligible: g pu o o co su g de ces input for work-consuming devices. one needs to know v as a function of P for the given process to perform the integration. compressors and pumps When the fluid is incompressible For the steady flow of a liquid through a device that involves no work interactions (such as a pipe section), the work term is zero (Bernoulli equation): It is developed for an internally reversible process and thus is applicable to incompressible fluids that involve no irreversibilities such as friction or shock waves The larger the specific volume, the greater the work produced (or consumed) by a steady-flow device. Reversible work relations for steady-flow and closed systems. 53
Discussion Note that compressing steam in the vapor form would require over 500 times more work than compressing it in the liquid form between the same pressure limits. Try to apply 1 st law for the pump and then for compressor 54
Example: Calculate the work produced, in kj/kg, for the reversible steady-flow process 1-3. Solution: 55
Example: Consider a steam power plant that operates between the pressure limits of 10 MPa and 20 kpa. Steam enters the pump as saturated liquid and leaves the turbine as saturated vapor. Determine the ratio of the work delivered by the turbine to the work consumed by the pump. Assume the entire cycle to be reversible and the heat losses from the pump and the turbine to be negligible. 56
Solution: Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are negligible. 3 The process is reversible. 4 The pump and the turbine are adiabatic. Properties The specific volume of saturated liquid water at 20 kpa is (Table A-5). Isentropic turbine: 57
The pump work input is determined from the steady-flow work relation to be You may apply the 1 st law for the pump, since you the state 2 is known from s 2 =s 1, and P 2 {w=(h 2 -h 1 )} Think about if the problem is not adiabatic and it is internally reversible!!! Is it reversible isothermal?? Is there a heat if s 1 =s 2, or s 3 =s 4??? 58
Proof that Steady-Flow Devices Deliver the Most and Consume the Least Work when the Process Is Reversible Taking heat input and work output positive: Actual Reversible Work-producing devices such as turbines deliver more work, and work- consuming devices such as pumps and compressors require less work when they operate reversibly. A reversible turbine delivers more work than an irreversible one if both operate between the same end states. 59
7-11 MINIMIZING THE COMPRESSOR WORK When kinetic and potential energies are negligible Three internally reversible cases for ideal gas(constant specific heats) : Isentropic (Pv k = constant): Steady-flow compression work Polytropic (Pv n = constant): The work input requirement for the polytropic case is decreases as the polytropic exponent n is decreased, by increasing the heat rejection during the compression process. If sufficient heat is removed, the value of n approaches unity and the process becomes isothermal. Isothermal (Pv = constant): The adiabatic compression (Pv k = constant) requires the maximum work and the isothermal compression (T = constant) requires the minimum. Why? P-v diagrams of isentropic, polytropic, and isothermal compression processes between the same pressure limits. One common way of cooling the gas during compression is to use cooling jackets around the casing of the compressors. 60
Obviously one way of minimizing the compressor work is to approximate an internally reversible process as much as possible by minimizing the irreversibilities such as friction, turbulence, and nonquasi-equilibrium compression. The extent to which this can be accomplished is limited by economic considerations. A second (and more practical) way of reducing the compressor work is to keep the specific volume of the gas as small as possible during the compression process. This is done by maintaining the temperature of the gas as low as possible during compression since the specific volume of a gas is proportional to temperature. t Therefore, reducing the work input to a compressor requires that t the gas be cooled as it is compressed. To have a better understanding of the effect of cooling during the compression process, we compare the work input requirements for three kinds of processes: 1- an isentropic process (involves no cooling) 2- a polytropic process (involves some cooling) 3- isothermal process (involves maximum cooling) Assuming all three processes are executed between the same pressure levels (P1 and P2) in an internally reversible manner and the gas behaves as an ideal gas (Pv = RT) with constant specific heats, 61
Multistage Compression with Intercooling The gas is compressed in stages and cooled between each stage by passing it through a heat exchanger called an intercooler. P-v and T-s diagrams for a twostage steady-flow compression process. The total work input for a two-stage compressor is the sum of the work inputs for each stage of compression: To minimize compression work during two-stage compression, the pressure ratio across each stage of the compressor must be the same. When this condition is satisfied, the compression work at each stage becomes identical, that is: 62
It is clear from these arguments that cooling a gas as it is compressed is desirable since this reduces the required work input to the compressor. However, often it is not possible to have adequate cooling through the casing of the compressor, and it becomes necessary to use other techniques to achieve effective cooling. One such technique is multistage compression with intercooling, where the gas is compressed in stages and cooled between each stage by passing it through a heat exchanger called an intercooler. Ideally, the cooling process takes place at constant pressure, and the gas is cooled to the initial temperature T 1 at each intercooler. Multistage compression with intercooling is especially attractive ti when a gas is to be compressed to very high h pressures. The gas is compressed in the first stage from P 1 to an intermediate pressure P x, cooled at constant pressure to the initial temperature T 1, and compressed in the second stage to the final pressure P 2. The compression processes, in general, can be modeled as polytropic (Pv n =constant) where the value of n varies between k and 1. 63
Example: Helium gas is compressed reversible from 100 kpa and 20 C to 850 kpa at a rate of 0.15 m 3 /s. Determine the power input to the compressor, assuming the compression process to be (a) isentropic, (b) polytropic with n = 1.2, (c) isothermal, and (d) ideal two-stage polytropic with n = 1.2. 64
Solution: Assume the Helium is an ideal gas with constant specific heats. R=2 2.0769 kpa.m 3 /kg.k., k = 1.667 (Table A-2). (a) Isentropic compression with k = 1.667: (b) Polytropic compression with n = 1.2: 65
(c) Isothermal compression: (d) Ideal two-stage compression with intercooling (n = 1.2): In this case, the pressure ratio across each stage is the same, and its value is determined from: The compressor work across each stage is also the same, thus total compressor work is twice the compression work for a single stage: Discussion Of all four cases considered, the isothermal compression requires the minimum work and the isentropic compression the maximum. The compressor work is decreased when two stages of polytropic compression are utilized instead of just one. As the number of compressor stages is increased, the compressor work approaches the value obtained for the isothermal case. 66
7-12 ISENTROPIC EFFICIENCIES OF STEADY-FLOW DEVICES The isentropic process involves no irreversibilities and serves as the ideal process for adiabatic devices. Isentropic Efficiency of Turbines Well-designed, large turbines have isentropic efficiencies above 90% For small turbines,however,, it may drop even below 70% The h-s diagram for the actual and isentropic processes of an adiabatic turbine. When kinetic and potential energies are negligible 67
Example: Steam enters an adiabatic turbine at 7 MPa, 600 C, and 80 m/s and leaves at 50 kpa, 150 C, and 140 m/s. If the power output of the turbine is 6 MW, determine (a) the mass flow rate of the steam flowing through the turbine and (b) the isentropic efficiency of the turbine. 68
Solution: Analysis (a) From the steam tables (Tables A-4 and A-6), The energy balance for this steady-flow system 69
(b) The isentropic exit enthalpy of the steam and the power output of the isentropic turbine are Discussion: if the KE and PE are negligible, you can use: 70
Isentropic Efficiencies of Compressors and Pumps Isothermal efficiency w t : reversible isothermal When kinetic and potential ti energies are negligible For a pump w s is a smaller quantity than w a, and this definition prevents from becoming greater than 100% The h-s diagram of the actual and isentropic processes of an adiabatic compressor. Compressors are sometimes intentionally cooled to minimize the work input. Can you use isentropic efficiency for a non-adiabatic compressor? Can you use isothermal efficiency for an adiabatic compressor? Well-designed compressors have isentropic efficiencies that range from 80 to 90% 71
Example: An adiabatic steady-flow device compresses argon at 200 kpa and 27 0 C to 2 MPa. If the argon leaves this compressor at 550 0 C, what is the isentropic efficiency of the compressor? Assume Argon is an ideal gas with constant specific heats Solution: k = 1.667 (Table A-2). The isentropic exit temperature is You can use the above equation to get the isentropic work since it is isentropic with constant specific heat, w comp, in = w s T 2 =T 2s, P 2 =P 2a =P 2s 72
Example: Air enters an adiabatic compressor at 100 kpa and 17 C atarateof2 a 2.4 m 3 /s, and it exits at 257 C C. The compressor has an isentropic efficiency of 84 percent. Neglecting the changes in kinetic and potential energies, determine (a) the exit pressure of air and (b) the power required to drive the compressor. Air is an ideal gas with variable specific heats. 73
Solution: R = 0.287 kpa.m 3 /kg.k (Table A-1) (a)) From the air table (Table A-17), Then from the isentropic relation 74
(b) The energy balance for this steady-flow system 75
Isentropic Efficiency of Nozzles Nozzles involve no work interactions, and the fluid experiences little or no change in its potential energy as it flows through the device. If, in addition, the inlet velocity of the fluid is small relative to the exit actual and velocity, the energy balance for this isentropic steady-flow device reduces to Then, The h-s diagram of the processes of an adiabatic nozzle. Isentropic efficiencies of nozzles are typically above 90% A substance leaves actual nozzles at a higher temperature (thus a lower velocity) as a result of friction. 76
Example: Air enters an adiabatic nozzle at 400 kpa and 547 C with low velocity and exits at 240 m/s. If the isentropic efficiency of the nozzle is 90 percent, determine the exit temperature and pressure of the air. Air is an ideal gas with variable specific heats Solution: From the air table (Table A-17), 77
From the air table we read 78
7-13 ENTROPY BALANCE the entropy change of a system during a process is equal to the net entropy transfer through the system boundary and the entropy generated within the system. Entropy Change of a System, S system Entropy change = Entropy at final state - Entropy at initial state Energy and =0.0 for steady-flow devices entropy balances When the properties of the for system are not uniform a system. The property entropy is a measure of molecular disorder or randomness of a system, and the second law of thermodynamics states that entropy can be created but it cannot be destroyed. d Therefore, the entropy change of a system during a process is greater than the entropy transfer by an amount equal to the entropy generated during the process within the system 79
Mechanisms of Entropy Transfer, S in and S out 1 Heat Transfer Heat transfer to a system increases the entropy of that system and thus the level of molecular disorder or randomness, and heat transfer from a system decreases it. In fact, heat rejection is the only way the entropy of a fixed mass can be decreased. d Entropy transfer by heat transfer: The quantity Q/T represents the entropy transfer accompanied by heat transfer, and the direction of entropy transfer is the same as the direction of heat transfer since thermodynamic temperature T is always a positive quantity. When the temperature T is not constant, Heat transfer is always accompanied by entropy transfer in the amount of Q/T, where T is the boundary temperature. where Q k is the heat transfer through the boundary at temperature T k at location k. When two systems are in contact, t the entropy transfer from the warmer system is equal to the entropy transfer into the cooler one at the point of contact. That is, no entropy can be created or destroyed at the boundary since the boundary has no thickness and occupies no volume. 80
Mechanisms of Entropy Transfer, S in and S out Entropy transfer by work: Note that work is entropy-free, and no entropy is transferred by work. Energy is transferred by both heat and work, whereas entropy is transferred only by heat. The first law of thermodynamics makes no distinction between heat transfer and work; it considers them as equals. The distinction between heat transfer and work is brought out by the second law: an energy interaction that is accompanied by entropy transfer is heat transfer, and an No entropy accompanies work as it energy interaction that is not accompanied crosses the system boundary. But by entropy transfer is work. That is, no entropy may be generated within the entropy is exchanged during a work system as work is dissipated into a interaction between a system and its less useful form of energy. surroundings. Thus, only energy is exchanged during work interaction whereas both energy and entropy are exchanged during heat transfer 81
Mechanisms of Entropy Transfer, S in and S out 2 Mass Flow Entropy transfer by mass: When the properties of the mass change during the process where A c is the cross-sectional area of the flow and V n is the local velocity normal to d A c Mass contains entropy as well as energy, and thus mass flow into or out of system is always accompanied by energy and entropy transfer. 82
Entropy Generation, S gen Entropy generation outside system boundaries can be accounted for by writing an entropy balance on an extended system that includes the system and its immediate surroundings. Fig. 7 61 Mechanisms of entropy transfer for a general system. Fig. 7 62 83
The term S gen represents the entropy generation within the system boundary only (Fig. 7 61), and not the entropy generation that t may occur outside the system boundary during the process as a result of external irreversibilities. Therefore, a process for which S gen = 0 is internally reversible, but not necessarily totally reversible. The total entropy generated during a process can be determined by applying the entropy balance to an extended system that includes the system itself and its immediate surroundings where external irreversibilities might be occurring (Fig. 7 62). Also, the entropy change in this case is equal to the sum of the entropy change of the system and the entropy change of the immediate surroundings. Note that under steady conditions, the state and thus the entropy of the immediate surroundings (let us call it the buffer zone ) at any point does not change during the process, and the entropy change of the buffer zone is zero. The entropy change of the buffer zone, if any, is usually small relative to the entropy change of the system, and thus it is usually disregarded. 84
Closed Systems Taking the positive direction of heat transfer to be to the system, the general entropy balance relation (Eq. 7 76) can be expressed for a closed system as Noting that any closed system and its surroundings can be treated as an adiabatic system and the total entropy change of a system is equal to the sum of the entropy changes of its parts, the entropy balance for a closed system and its surroundings can be written as if its temperature is constant. 85
The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. Control Volumes Taking the positive direction of heat transfer to be to the system, Eq. 6-76 The entropy of a control volume changes as a result of mass flow as well as heat transfer. The rate of entropy change within the control volume during a process is equal to the sum of the rate of entropy transfer through the control volume boundary by heat transfer, the net rate of entropy transfer into the control volume by mass flow, and the rate of entropy generation within the boundaries of the control volume as a result of irreversibilities. The entropy of a substance always increases (or remains constant in the case of a reversible process) as it flows through a single-stream, adiabatic, steady-flow device. 86
EXAMPLE 7-17 Consider steady heat transfer through a 5-m x 7-m brick wall of a house of 30 cm. On a day when the temperature of the outdoors is 0 o C, the house is maintained at 27 o C. The temperatures of the inner and outer surfaces of the brick wall are measured to be 20 o C and 5 o C, respectively, and the rate of heat transfer through the wall is 1035 W. Determine the rate of entropy generation in the wall, and the rate of total entropy generation associated with this heat transfer process. 87
Solution: This is a closed system since no mass crosses the system boundary during the process. The rate form of the entropy balance for the wall simplifies to Therefore, the rate of entropy generation in the wall is You can work on isolated system: S gen =ΔS sys + ΔS surr, L + ΔS surr, R = -1035/(27+273)+ 1035/(0+273)=0.341 Remember: [Thermal Reservoir]Internally reversible isothermal process : Q=T 0 ΔS To determine the rate of total entropy generation during this heat transfer process, we extend the system to include the regions on both sides of the wall that experience a temperature change. The entropy balance for this extended system (system + immediate surroundings) 0 The differences between the two entropy generations is 0.150 W/K, and it represents the entropy generated in the air layers on both sides of the wall. The entropy generation in this case is entirely due to irreversible heat transfer through a finite temperature difference. 88
Example: Refrigerant-134a is throttled from 900 kpa and 35 C to 200 kpa. Heat is lost from the refrigerant in the amount of 0.8 kj/kg to the surroundings at 25 C. Determine (a) the exit temp of the refrigerant and (b) the entropy generation during this process. Solution: Use the extended system because of heat transfer, if no heat transfer use the system boundary: Energy Balance s 1 -s 2 -q/t+s gen =0.00 0.37135-0.38884-0.8/(25+273)+s gen =0.0 0 Entropy Balance s gen = 0.0202 kj/kg.k 89
Example: Air (c p = 1.005 kj/kg C) is to be preheated by hot exhaust gases in a cross-flow heat exchanger before it enters the furnace. Air enters the heat exchanger at 95 kpa and 20 C at a rate of 1.6 m 3 /s. The combustion gases (c p = 1.10 kj/kg C) enter at 180 C atarateof2.2 a kg/s and leave at 95 C. Determine (a) the rate of heat transfer to the air, (b) the outlet temperature of the air, and (c) the rate of entropy generation. 90
Solution: Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. R air = 0.287 kj/kg.k (Table A-1) We take the exhaust pipes as the system, which is a control volume Then the rate of heat transfer from the exhaust gases becomes The mass flow rate of air is 91
Noting that heat loss by the exhaust gases is equal to the heat gain by the air, the outlet temperature of the air becomes The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger: Discussion: if the fluids are saturated liquid as an example, use the Tables 92
Example: An ordinary egg can be approximated as a 5.5- cm diameter sphere. The egg is initially at a uniform temp of 8 C and is dropped into boiling water at 97 C. Taking the properties of the egg to be ρ=1020 kg/m 3 and c p =3.32 kj/kg C, determine (a) how much heat is transferred to the egg by the time the average temperature of the egg rises to 70 C and (b) the amount of entropy generation associated with this heat transfer process. 93
Solution We take the egg as the system. This is a closes system since no mass enters or leaves the egg. The energy balance for this closed system can be expressed as We can use an isolated system: S gen = ΔS system + Δ S surr =0.0588-18.3/(97+273) =0.00934 kj/kg The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the egg and its immediate surroundings We could not use the closed system(egg) alone to calculate S gen because the boundary Temp is not constant 94
Example: Steam expands in a turbine steadily at a rate of 25,000 kg/h, entering at 6 MPa and 450 C and leaving at 20 kpa as saturated vapor. If the power generated by the turbine is 4 MW, determine the rate of entropy generation for this process. Assume the surrounding medium is at 25 C C. 95
Solution applying the entropy balance on an extended system that includes the turbine and its immediate surroundings so that the boundary temperature of the extended system is 25 C at all times. Note: in steady-device device take the system boundary if no heat, while take extended system if any heat 96
Example: Liquid water at 200 kpa and 20 C is heated in a chamber by mixing it with superheated steam at 200 kpa and 150 C. Liquid water enters the mixing chamber at a rate of 2.5 kg/s, and the chamber is estimated to lose heat to the surrounding air at 25 C at a rate of 1200 kj/min. If the mixture leaves the mixing chamber at 200 kpa and 60 C, determine (a) the mass flow rate of the superheated steam and (b) the rate of entropy generation during this mixing process. 97
Solution 98
(b) The rate of total entropy generation during this process is determined by applying the entropy balance on an extended system that includes the mixing chamber and its immediate surroundings so that the boundary temperature t of the extended d system is 25 C at all times. It gives 99
Example: A 0.3-m 3 rigid tank is filled with saturated liquid id water at 150 C. A valve at the bottom of the tank is now opened, and one-half of the total mass is withdrawn from the tank in the liquid form. Heat is transferred to water from a source at 200 C so that the temperature in the tank remains constant. Determine (a) the amount of heat transfer and (b) the total entropy generation for this process. 100
Solution The direction of heat transfer is to the tank (will be verified). The properties of water are (Tables A-4 through A-6) 101
(b) The total entropy generation is determined by considering a combined system that includes the tank and the heat source. Noting that no heat crosses the boundaries of this combined system and no mass enters, the entropy balance for it can be expressed as 102
Example: An iron block of unknown mass at 85 C is dropped into an insulated tank that contains 20 Lof water aterat20 C 20 C. Atthe the same time, a paddle wheel driven by a 200-W motor is activated to stir the water. Thermal equilibrium is established after 10 min with a final temperature t of 25 C. Determine (a) the mass of the iron block and (b) the entropy generated during this process. 103